Second-order operators. The Laplacian

Repeated applications of the first order operators gradient, divergence and curl yields higher-order operators. If we limit ourselves to second derivatives, then nine different combinations of the three first-order operators are possible, but four of them do not make sense (remember that operators are applied sequentially from right to left):

$\nabla \nabla$	:	the gradient of a scalar is a vector, to which the gradient cannot be applied;
$\nabla \times \nabla$	:	the curl of a vector is a vector, to which the gradient cannot be applied;
$\nabla \cdot \nabla \cdot$	:	the divergence of a vector is a scalar, to which the divergence cannot be applied;
$\nabla \times \nabla \cdot$	:	the divergence of a vector is a scalar, to which the curl cannot be applied.

Of the remaining five second-order operators, two are identically zero:

$$\boxed{ \nabla \cdot \nabla \times \equiv 0 \space , \nabla \times \nabla \equiv 0 }$$ (2.11)

The proof of these two identities is left as an exercise (see Problems). There remain three second-order operators, that are linked by the following relation, whose proof is also left as an exercise (see Problems):

$$\boxed{ \nabla \times \nabla \times = \nabla \, \nabla \cdot \, - \, \nabla \cdot \nabla}$$ (2.12)

here only two of the three vector quantities in this last equation are independent of each other. At first, the last equation seems to make no sense, because $\nabla \times \nabla \times$ and $\nabla \nabla \cdot$ apply to vectors, whereas $\nabla \cdot \nabla$ applies to a scalar. The operator

$$\boxed{ \nabla \cdot \nabla = \nabla^2 =\frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} + \frac{\partial^2}{\partial z^2} }$$ (2.13)

is called the laplacian (from the French mathematician and astronomer Laplace). The laplacian applies to a scalar function of position; however, by convention, the laplacian of a vector is meant to be a vector whose components are the laplacians of the components of the original vector, in rectangular coordinates, that is:

$$\nabla^2 \underline{F} \triangleq \hat{x} \nabla^2 F_x + \hat{y} \nabla^2 F_y + \hat{z} \nabla^2 F_z \space .$$ (2.14)

Note that although the laplacians of $F_x$, $F_y$, and $F_z$ in the last formula may be calculated in any coordinate system, the vector $\underline{F}$ must be split in rectangular components for (A.12) to be valid.